A157697 Decimal expansion of sqrt(2/3).
8, 1, 6, 4, 9, 6, 5, 8, 0, 9, 2, 7, 7, 2, 6, 0, 3, 2, 7, 3, 2, 4, 2, 8, 0, 2, 4, 9, 0, 1, 9, 6, 3, 7, 9, 7, 3, 2, 1, 9, 8, 2, 4, 9, 3, 5, 5, 2, 2, 2, 3, 3, 7, 6, 1, 4, 4, 2, 3, 0, 8, 5, 5, 7, 5, 0, 3, 2, 0, 1, 2, 5, 8, 1, 9, 1, 0, 5, 0, 0, 8, 8, 4, 6, 6, 1, 9, 8, 1, 1, 0, 3, 4, 8, 8, 0, 0, 7, 8, 2, 7, 2, 8, 6, 4
Offset: 0
Examples
0.81649658092772603273242802490196379732198249355222...
References
- L. B. W. Jolley, Summation of Series, Dover, 1961, eq. (168) on page 32.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..10000
- Jisho Kotani, Problem 2053, Crux Mathematicorum, Vol. 5, No. 1 (1995), p. 202; Solution to Problem 2053, by David Hankin, ibid., Vol. 22, No. 4 (1996), pp. 187-188.
- D. H. Lehmer, Interesting series involving the Central Binomial Coefficient, Am. Math. Monthly, Vol. 92, No. 7 (1985), pp. 449-457.
- Index entries for algebraic numbers, degree 2.
Programs
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Magma
Sqrt(2/3); // G. C. Greubel, Mar 30 2018
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Maple
evalf(sqrt(2/3)) ;
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Mathematica
RealDigits[Sqrt[2/3], 10, 200][[1]] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011*)
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PARI
sqrt(2/3) \\ G. C. Greubel, Mar 30 2018
Formula
Equals 1 - (1/2)/2 + (1*3)/(2*4)/2^2 - (1*3*5)/(2*4*6)/2^3 + ... [Jolley]
Equals Sum_{n>=0} (-1)^n*binomial(2n,n)/8^n = 1/A115754. Averaging this constant with sqrt(2) = A002193 = Sum_{n>=0} binomial(2n,n)/8^n yields A145439.
From Michal Paulovic, Dec 08 2022: (Start)
Equals 2 * A020763.
Has periodic continued fraction expansion [0, 1, 4; 2, 4]. (End)
Equals exp(-arctanh(1/5)). - Amiram Eldar, Jul 10 2023
Equals Product_{k>=1} (1 + (-1)^k/A092259(k)). - Amiram Eldar, Nov 24 2024
Comments